\documentclass[]{report}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{venndiagram}
\usepackage{float}
\usepackage{amsthm}

\title{Introductory Mathematics: Algebra and Analysis Solutions}
\author{Michael Rocke}

\begin{document}

\maketitle

\section{Chapter 1}

\subsection{Exercises}
Notes \\

$\mathbb{N} $ = Set of Natural numbers, $ \{1, 2, 3, .. \}$ \\

$\mathbb{Z} $ = Set of Integers, $ \{..., -2, -1, 0, 1, 2, .. \}$ \\

$\mathbb{Q} $ = Set of Rational Numbers, $ Q = \{\frac{a}{b} | a,b \in \mathbb{Z}, b \ne 0 \} $ \\

$\mathbb{R} $ = Set of Real numbers

\subsubsection{1.1}
$ A = \{1,2,3\} , B = \{1, 2\}, C=\{1, 3\}, D=\{2, 3\}, E=\{1\}, F=\{2\}, G=\{3\}, H=\emptyset$ \\


a) $A \cap B = B $ \\

b) $A \cup C = A $ \\

c) $A \cap (B \cap C) = E $ \\

d) $(C \cup A) \cap B  = B $ \\

e) $ A \setminus B = G $ \\

f) $ C \setminus A = H $ \\

g) $ (D \setminus F) \cup (F \setminus D)  = G$ \\

h) $ G \setminus A = H$ \\

j) $ A \cup ((B \setminus C) \setminus F)  = A$\\

k) $ H \cup H = H $\\

l) $ A \cap A = A $\\

m) $ ((B \cup C) \cap C) \cup H = C $\\

\subsubsection{1.2}

a) i and ii are the same, iii is different \\
b) i and ii are the same, iii is different \\
c) $ i = \{1, 2, 3, 4, 5, 6, 7\}, ii = \{1, 2, 3, 4, 5, 6, 7, -1, -2, -3, -4, -5, -6, -7\}, iii = \{1, 2, 3, 4, 5, 6, 7\} $, so i and iii are the same, ii is different \\
d) $ i = \{0, 1, 2, 3, ...\}, ii = \{1, 2, 3, ...\}, iii = \{1, 2, 3, ...\}$, ii and iii are the same, i is different \\
e) i and iii are the same, ii is different \\
\\
f) ii and iii are same, i is different \\
\\
g) ii and iii are same, i is different \\
h) i and iii are same, ii is different \\
\\
j) $ i = \emptyset, ii = \emptyset, iii = \{\emptyset\} $ i and ii are same, iii are different \\
k) ii and iii are the same, i is different \\
l) ii and iii are the same, i is different \\
m) $ i = \{\emptyset, \{\emptyset\}, 0\}, ii = \{\emptyset, \{\emptyset\}, 0\}, iii  = \{\emptyset, 0\} $ i and ii are same, iii different\\ 

\subsubsection{1.3}


a)


\begin{figure} [H]
	\begin{venndiagram3sets}
		\fillACapBCapC
	\end{venndiagram3sets}
	\caption{ $ A \cap B \cap C $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillACapBNotC
	\end{venndiagram3sets}
	\caption{ $ A \cap B \cap C' $}
\end{figure}




\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillACapCNotB
	\end{venndiagram3sets}
	\caption{ $ A \cap B' \cap C $}
\end{figure}




\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillBCapCNotA
	\end{venndiagram3sets}
	\caption{ $ A' \cap B \cap C $}
\end{figure}


\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillOnlyA
	\end{venndiagram3sets}
	\caption{ $ A \cap B' \cap C' $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillOnlyB
	\end{venndiagram3sets}
	\caption{ $ A' \cap B \cap C' $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillOnlyC
	\end{venndiagram3sets}
	\caption{ $ A' \cap B' \cap C $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillNotABC
	\end{venndiagram3sets}
	\caption{ $ A' \cap B' \cap C' $}
\end{figure}

b) 



\begin{figure} [H]
	\begin{venndiagram3sets}
		\fillACapBCapC
	\end{venndiagram3sets}
	\caption{ $ $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillACapBNotC
	\end{venndiagram3sets}
	\caption{ $  $}
\end{figure}




\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillACapCNotB
	\end{venndiagram3sets}
	\caption{ $ $}
\end{figure}




\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillBCapCNotA
	\end{venndiagram3sets}
	\caption{ $ $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillOnlyA
	\end{venndiagram3sets}
	\caption{ $ (A' \cup B \cup C)' $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillOnlyB
	\end{venndiagram3sets}
	\caption{ $ (A \cup B' \cup C)' $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillOnlyC
	\end{venndiagram3sets}
	\caption{ $ (A \cup B \cup C')' $}
\end{figure}



\begin{figure}[H]
	\begin{venndiagram3sets}
		\fillNotABC
	\end{venndiagram3sets}
	\caption{ $ (A \cup B \cup C)' $}
\end{figure}


\subsubsection{1.4}

a) Prove that ${1, 2, 3} = {1, 1, 2, 3}$

\begin{proof}
Lets define the following sets
$A = \{1, 2, 3\}$ and $B = \{1, 1, 2, 3\}$. For two sets to be equal, every element in one set must belong to the other, meeting the property $A = A \cup B = B$.
Just looking empirically, $A$ contains the elements $1, 2, 3$ and B contains the same elements. 
We can also prove this by defining both sets as follows
$A = \{a | a \in \mathbb{N},  1 \leq a \leq 3\}$ and $B = \{b | b \in \mathbb{N},  1 \leq b \leq 3\}$ which is the same definition
\end{proof}

b) Using $P(A)$ to represent the Power Set of $A$. How many elements are there in the set $P(\{1,2,3\})$? $2^3 = 8$. These are: $\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}$

c)  Deterime the cardinality of $P(\emptyset)$? $2^0 = 1$ resulting in $\{\emptyset\}$

d) $P(P(\emptyset))$? $2^1 = 2$ $\{\emptyset, \{\emptyset\}\}$

e)  $P(P(P(P(P(P(\emptyset)))))))$? Using substitution, we can see $P(P(P(P(\{\emptyset, \{\emptyset\}\}))))$, and the cardinality of a powerset is defined $2^n$ where  $n$ is the cardinality of the  argument set.  This becomes the following

\begin{align*}
A_1 = P(\emptyset), |A_1| = 2^0 = 1\\
A_2 = P(A_1), |A_2| = 2^1 = 2 \\
A_4 = P(A_2), |A_4| = 2^2 = 4 \\
A_16 = P(A_4), |A_ 16| = 2^4 = 16 \\
A_65536 = P(A_16), |A_ 65536| = 2^16 = 65536
\end{align*}

\subsubsection{1.5}

a) Draw Venn diagrams illustrating the truth to De Morgan's laws

$(A \cup B)' = A' \cap  B'$

\begin{figure}[H]
	\begin{venndiagram2sets}
		\fillNotAorB
	\end{venndiagram2sets}
= \\
	\begin{venndiagram2sets}
		\fillNotA
	\end{venndiagram2sets}
    $\cap$
    \begin{venndiagram2sets}
    	\fillNotB
    \end{venndiagram2sets}
= \\
(    \begin{venndiagram2sets}
    	\fillA
    \end{venndiagram2sets}
$\cup$
    \begin{venndiagram2sets}
    	\fillB
    \end{venndiagram2sets})'
\end{figure}

$(A \cap B)' = A' \cup B'$

\begin{figure}[H]
	\begin{venndiagram2sets}
		\fillNotAorNotB
	\end{venndiagram2sets}
	= \\
	\begin{venndiagram2sets}
		\fillNotA
	\end{venndiagram2sets}
	$\cup$
	\begin{venndiagram2sets}
		\fillNotB
	\end{venndiagram2sets}
	= \\
	(    \begin{venndiagram2sets}
		\fillA
	\end{venndiagram2sets}
	$\cap$
	\begin{venndiagram2sets}
		\fillB
	\end{venndiagram2sets})'
\end{figure}

b) Prove De Morgan's laws.

\begin{proof}
	First we prove $(A \cup B)' = A' \cap B'$. Lets denote the set $X = (A \cup B)'$ and $Y = A' \cap B'$. In English, $X$ contains all the items that aren't in either A or B. That would state that $A$ is not a subset of $X$ and same goes for $B$, e.g. $X = U - (A + B)$ where U is the universal set. $Y$ in English states that it contains all the elements which are not in $A$ and not in $B$, e.g. $Y= U - A -  B$, which we can see that $Y = X$
\end{proof}
\end{document}